Question: Find the center of the circle with equation $9x^2-18x+9y^2+36y+44=0.$
Explanation: First, we factor out the constants of the squared terms to get $9(x^2-2x)+9(y^2+4y)=-44$.

To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$ and $\left(\dfrac{4}{2}\right)^2=4$ after the $4y$, giving $9(x-1)^2+9(y+2)^2=-44+9+36=1$. Dividing the equation by $9$ gives $(x-1)^2+(y+2)^2=\dfrac{1}{9}$, so the center is $\boxed{(1,-2)}$.